我在网上发现了一个很有挑战性的编码问题，我想我会试试。
一般的想法是，给定文本字符串T和模式P，找到该模式的出现，将其相应的值相加，返回max和min。如果您想更详细地阅读该问题，请参阅this。
然而，下面是我提供的代码，它适用于一个简单的测试用例，但是当在多个复杂的测试用例上运行时，它非常慢，而且我不确定我的代码需要优化到哪里。
有谁能帮我把逻辑搞错吗？
public class DeterminingDNAHealth {
private DeterminingDNAHealth() {
/*
* Fixme:
* Each DNA contains number of genes
* - some of them are beneficial and increase DNA's total health
* - Each Gene has a health value
* ======
* - Total health of DNA = sum of all health values of beneficial genes
*/
}
int checking(int start, int end, String pattern) {
String[] genesChar = new String[] {
"a",
"b",
"c",
"aa",
"d",
"b"
};
String numbers = "123456";
int total = 0;
for (int i = start; i <= end; i++) {
total += KMPAlgorithm.initiateAlgorithm(pattern, genesChar[i]) * (i + 1);
}
return total;
}
public static void main(String[] args) {
String[] genesChar = new String[] {
"a",
"b",
"c",
"aa",
"d",
"b"
};
Gene[] genes = new Gene[genesChar.length];
for (int i = 0; i < 6; i++) {
genes[i] = new Gene(genesChar[i], i + 1);
}
String[] checking = "15caaab 04xyz 24bcdybc".split(" ");
DeterminingDNAHealth DNA = new DeterminingDNAHealth();
int i, mostHealthiest, mostUnhealthiest;
mostHealthiest = Integer.MIN_VALUE;
mostUnhealthiest = Integer.MAX_VALUE;
for (i = 0; i < checking.length; i++) {
int start = Character.getNumericValue(checking[i].charAt(0));
int end = Character.getNumericValue(checking[i].charAt(1));
String pattern = checking[i].substring(2, checking[i].length());
int check = DNA.checking(start, end, pattern);
if (check > mostHealthiest)
mostHealthiest = check;
else
if (check < mostUnhealthiest)
mostUnhealthiest = check;
}
System.out.println(mostHealthiest + " " + mostUnhealthiest);
// DNA.checking(1,5, "caaab");
}
}
KMPAlgorithm公司
public class KMPAlgorithm {
KMPAlgorithm() {}
public static int initiateAlgorithm(String text, String pattern) {
// let us generate our LPC table from the pattern
int[] partialMatchTable = partialMatchTable(pattern);
int matchedOccurrences = 0;
// initially we don't have anything matched, so 0
int partialMatchLength = 0;
// we then start to loop through the text, !note, not the pattern. The text that we are testing the pattern on
for (int i = 0; i < text.length(); i++) {
// if there is a mismatch and there's no previous match, then we've hit the base-case, hence break from while{...}
while (partialMatchLength > 0 && text.charAt(i) != pattern.charAt(partialMatchLength)) {
/*
* otherwise, based on the number of chars matched, we decrement it by 1.
* In fact, this is the unique part of this algorithm. It is this part that we plan to skip partialMatchLength
* iterations. So if our partialMatchLength was 5, then we are going to skip (5 - 1) iteration.
*/
partialMatchLength = partialMatchTable[partialMatchLength - 1];
}
// if however we have a char that matches the current text[i]
if (text.charAt(i) == pattern.charAt(partialMatchLength)) {
// then increment position, so hence we check the next char of the pattern against the next char in text
partialMatchLength++;
// we will know that we're at the end of the pattern matching, if the matched length is same as the pattern length
if (partialMatchLength == pattern.length()) {
// to get the starting index of the matched pattern in text, apply this formula (i - (partialMatchLength - 1))
// this line increments when a match string occurs multiple times;
matchedOccurrences++;
// just before when we have a full matched pattern, we want to test for multiple occurrences, so we make
// our match length incomplete, and let it run longer.
partialMatchLength = partialMatchTable[partialMatchLength - 1];
}
}
}
return matchedOccurrences;
}
private static int[] partialMatchTable(String pattern) {
/*
* TODO
* Note:
* => Proper prefix: All the characters in a string, with one or more cut off the end.
* => proper suffix: All the characters in a string, with one or more cut off the beginning.
*
* 1.) Take the pattern and construct a partial match table
*
* To construct partial match table {
* 1. Loop through the String(pattern)
* 2. Create a table of size String(pattern).length
* 3. For each character c[i], get The length of the longest proper prefix in the (sub)pattern
* that matches a proper suffix in the same (sub)pattern
* }
*/
// we will need two incremental variables
int i, j;
// an LSP table also known as “longest suffix-prefix”
int[] LSP = new int[pattern.length()];
// our initial case is that the first element is set to 0
LSP[0] = 0;
// loop through the pattern...
for (i = 1; i < pattern.length(); i++) {
// set our j as previous elements data (not the index)
j = LSP[i - 1];
// we will be comparing previous and current elements data. ei char
char current = pattern.charAt(i), previous = pattern.charAt(j);
// we will have a case when we're somewhere in loop and two chars will not match, and j is not in base case.
while (j > 0 && current != previous)
// we decrement our j
j = LSP[j - 1];
// simply put, if two characters are same, then we update our LSP to say that at that point, we hold the j's value
if (current == previous)
// increment our j
j++;
// update the table
LSP[i] = j;
}
return LSP;
}
}
将代码贷记到Github
最佳答案
您可以尝试这个KMP实现它是o（m+n），正如kmp所期望的那样。应该快得多：
private static int[] failureFunction(char[] pattern) {
int m = pattern.length;
int[] f = new int[pattern.length];
f[0] = 0;
int i = 1;
int j = 0;
while (i < m) {
if (pattern[i] == pattern[j]) {
f[i] = j + 1;
i++;
j++;
} else if (j > 0) {
j = f[j - 1];
} else {
f[i] = 0;
i++;
}
}
return f;
}
private static int kmpMatch(char[] text, char[] pattern) {
int[] f = failureFunction(pattern);
int m = pattern.length;
int n = text.length;
int i = 0;
int j = 0;
while (i < n) {
if (pattern[j] == text[i]) {
if (j == m - 1){
return i - (m - 1);
} else {
i++;
j++;
}
} else if (j > 0) {
j = f[j - 1];
} else {
i++;
}
}
return -1;
}
关于java - KMP字符串搜索算法？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/46821536/